UVALive 7037 (最大密度子图 网络流)

每个数都看成一个节点,每个逆序对之间的节点连边,于是只需要求出这个图的最大密度子图,可以用最小割模型解决.

#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>#include <vector>#include <queue>using namespace std;const double INF = 1e8;#define maxn 1511#define maxm 2111111#define eps 1e-10#define type doubleint n, m, x, N;int s, t;struct Edge{    int from, to,next;    type cap,flow;    void get(int u,int a,int b,type c,type d)    {        from = u; to = a; next = b; cap = c; flow = d;    }}edge[maxm];int tol;int head[maxn];int gap[maxn],dep[maxn],pre[maxn],cur[maxn];void init(){    tol=0;    memset(head,-1,sizeof(head));}void add_edge(int u,int v,type w,type rw=0){ //cout << u << " " << v << " " << w << endl;    edge[tol].get(u, v,head[u],w,0);head[u]=tol++;    edge[tol].get(v, u,head[v],rw,0);head[v]=tol++;}type sap(int start,int end,int N){    memset(gap,0,sizeof(gap));    memset(dep,0,sizeof(dep));    memcpy(cur,head,sizeof(head));    int u=start;    pre[u]=-1;    gap[0]=N;    type ans=0;    while(dep[start]<N)    {        if(u==end)        {            type Min=INF;            for(int i=pre[u];i!=-1;i=pre[edge[i^1].to])                if(Min>edge[i].cap-edge[i].flow)                   Min=edge[i].cap-edge[i].flow;            for(int i=pre[u];i!=-1;i=pre[edge[i^1].to])            {                edge[i].flow+=Min;                edge[i^1].flow-=Min;            }            u = start;            ans+=Min;            continue;        }        bool flag=false;        int v;        for(int i=cur[u];i !=-1;i=edge[i].next)        {            v=edge[i].to;            if(edge[i].cap-edge[i].flow&&dep[v]+1==dep[u])            {                flag=true;                cur[u]=pre[v]=i;                break;            }        }        if(flag)        {            u=v;            continue;        }        int Min=N;        for(int i=head[u];i!=-1;i=edge[i].next)            if(edge[i].cap-edge[i].flow&&dep[edge[i].to]<Min)        {            Min=dep[edge[i].to];            cur[u]=i;        }        gap[dep[u]]--;        if(!gap[dep[u]]) return ans;        dep[u]=Min+1;        gap[dep[u]]++;        if(u!=start) u=edge[pre[u]^1].to;    }    return ans;}int degree[maxn];int num[maxn];type f (type g) {     init ();    s = 0, t = n+1, N = n+2;    double U = 1.0*m;    for (int i = 1; i <= n; i++) {        for (int j = 1; j < i; j++) {            if (num[j] > num[i]) {                add_edge (i, j, 1);                add_edge (j, i, 1);            }        }    }    for (int i = 1; i <= n; i++)        add_edge (s, i, U);    for (int i = 1; i <= n; i++)        add_edge (i, t, U+2*g-1.0*degree[i]);    double ans = sap (s, t, N);     ans = (U*n-ans)/2;    return ans;}bool vis[maxn];void dfs (int u) {    vis[u] = 1;    for (int i = head[u]; i != -1; i = edge[i].next) {        int v = edge[i].to;        if (vis[v])            continue;        if (edge[i].cap-edge[i].flow > 0 && edge[i].cap > 0)            dfs (v);    }}int main () {    //freopen ("in.txt", "r", stdin);    int tt, kase = 0;    scanf ("%d", &tt);    while (tt--) {        scanf ("%d", &n);        m = 0;        memset (degree, 0, sizeof degree);        for (int i = 1; i <= n; i++) {            scanf ("%d", &num[i]);            for (int j = 1; j < i; j++) {                if (num[j] > num[i]) {                    degree[i]++;                    degree[j]++;                    m++;                }            }        }        double l = 0, r = 10000;        int kk = 50;        while (kk--) {            double mid = (l+r)/2;            double h = f (mid);            if (h <= 0)                r = mid;            else                l = mid;        }        printf ("Case #%d: %.7f\n", ++kase, l);    }    return 0;}